A certain metal when irradiated to light (v=3.2×10^16 Hz) emits photoe...
A certain metal when irradiated to light (v=3.2×10^16 Hz) emits photoe...
Introduction:
When a metal is irradiated with light, it can emit photoelectrons. The kinetic energy of these photoelectrons depends on the frequency (v) of the incident light. In this scenario, we are given two different frequencies of light, and we need to determine the threshold frequency of the metal.
Threshold Frequency:
The threshold frequency (v0) is the minimum frequency of light required to remove an electron from the metal's surface. Below this frequency, no photoelectrons are emitted. The energy of a photon (E) can be calculated using the equation: E = hv, where h is the Planck's constant (6.626 x 10^-34 J.s).
Photoelectrons Comparison:
According to the given information, when the metal is irradiated with light of frequency v = 2.0 x 10^16 Hz, the photoelectrons have a certain kinetic energy (K1). When the metal is irradiated with light of frequency v = 3.2 x 10^16 Hz, the photoelectrons have twice the kinetic energy (2K1).
Explanation:
We can use the equation for the kinetic energy of a photoelectron to compare the two cases:
K1 = hv1 - φ (1)
2K1 = hv2 - φ (2)
where φ is the work function of the metal, representing the minimum energy required to remove an electron from its surface.
Calculations:
Substituting equation (1) into equation (2), we get:
hv2 - φ = 2(hv1 - φ)
hv2 - φ = 2hv1 - 2φ
Rearranging the equation, we have:
hv2 - 2hv1 = φ
Since we are interested in the threshold frequency (v0), we can set the kinetic energy of the photoelectron to zero:
0 = hv0 - φ
Simplifying the equation, we get:
hv0 = φ
Conclusion:
From the equation hv0 = φ, we can conclude that the threshold frequency (v0) is equal to the ratio of the work function (φ) to Planck's constant (h). Therefore, the threshold frequency of the metal is given by option (c) 1.2 x 10^16 Hz.
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